HDU 6069 Counting Divisors —— 2017 Multi-University Training 4

Counting Divisors

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 2599    Accepted Submission(s): 959

Problem Description

In mathematics, the function d(n) denotes the number of divisors of positive integer n.

For example, d(12)=6 because 1,2,3,4,6,12 are all 12's divisors.

In this problem, given l,r and k, your task is to calculate the following thing :

 

Input

The first line of the input contains an integer T(1≤T≤15), denoting the number of test cases.

In each test case, there are 3 integers l,r,k(1≤l≤r≤1012,r−l≤106,1≤k≤107).

 

Output

For each test case, print a single line containing an integer, denoting the answer.

 

Sample Input

3

1 5 1

1 10 2

1 100 3

 

Sample Output

10

48

2302

 

题目大意：d（i）是 i 的因数个数，让我们求 l<=i<=r 时，d（i^k）之和.

思路：对一个数n=p₁^t₁*p₂^t₂*...*p_n^t_n, p_i是n的质因数。则n的因数个数是(t₁+1)*(t₂+1)*...*(t_n-1+1)*(t_n+1), 易得i^k的因数个数是(k*t₁+1)*(k*t₂+1)*...*(k*t_n+1)，那么接下来就是要对 i 进行质因数分解了。 在打表打出质因数后，分解时对于一个质数P, 在[l , r]区间内所有能整除P的数进行质因数分解，这样能保证不会有多余时间花在搜索质因数上，这种做法类似筛法。具体见代码。

 

AC代码(标程)：

 #include<iostream>
 #include<cstring>
 #include<cstdio>
 #include<cmath>
 #include<stdio.h>
 #define it (p-l)
 using namespace std;
 typedef long long LL;
 const LL MOD=;
 const long long MAXN=;
 long long prime[MAXN],tot=;
 bool isPrime[MAXN];
 LL k,num[MAXN],res[MAXN];
 void getprime(){
     memset(isPrime, true, sizeof(isPrime));
     for(int i=;i<MAXN;i++){
         if(isPrime[i]){
             prime[++tot]=i;
         }
         for(int j=;j<=tot;j++){
             if(i*prime[j]>MAXN) break;
                 isPrime[i*prime[j]]=false;
             if(i%prime[j]==) break;
         }
     }
     return ;
 }
 LL cal(LL l, LL r)
 {
     LL ans=,tmp,cnt;
     for(int i=;i<=tot;i++)
     {
         LL p=(l+prime[i]-)/prime[i]*prime[i];
         while(p<=r){
             cnt=;
             while(num[it]%prime[i]==){
                 num[it]/=prime[i];
                 cnt++;
             }
             res[it]=res[it]*(k*cnt+)%MOD;
             p+=prime[i];
         }
     }
     for(LL p=l;p<=r;p++){
         if(num[it]==)
             ans+=res[it];
         else
             ans+=res[it]*(k+);
         ans%=MOD;
     }
     return ans;
 }
 int main()
 {
     int T;
     LL l,r;
     getprime();
     scanf("%d", &T);
     while(T--)
     {
         scanf("%lld %lld %lld", &l, &r, &k);
         for(LL p=l;p<=r;p++){
             res[it]=;
             num[it]=p;
         }
         LL res=cal(l, r);
         printf("%lld\n", res);
     }
 }